In a city of western country, 70% married people take divorce. What is the probability that atleast 3 among 4 people will take divorce?

• Let,married people take divorce(p)=0.7 and married people not take divorce(q)=0.3
  so,probability of atleast 3 among 4 people take divorce=4C3*p^3*q+4C4*p^4*q^0
  = 4*0.343*0.3+1*0.2401
  = 0.4116+0.2401
  = 0.6517(Ans.)
• 11 years agoHelpfull: Yes(10) No(3)
• We have to, basically find d probability of 75 people being divorced in a population of 100. Amongst 92 people, 69 can take divorce and in rest 8, 1 may take divorce.
  Therefore, our condition is satisfied for 92/4 = 23 cases.
  Its not satisfied for 8/4 = 2 cases.
  Hence, probability = 23/25 = 0.92
• 11 years agoHelpfull: Yes(5) No(1)
• 70/75=14/15
• 11 years agoHelpfull: Yes(1) No(0)